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Physics - Rotation Question with Solution | TestHub

PhysicsRotationCalculation of M.I.Medium2 minPYQ_2016

PhysicsMediumsingle choice

From a disc of radius $R$ and mass $M$ , a circular hole of diameter $R$ , whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through the centre?NEET 2016 (Phase 1)

Options:

Answer:

Solution:

$I = \frac{M R^{2}}{2} - \frac{3 σ}{2} π {(\frac{R}{2})}^{2} {(\frac{R}{2})}^{2}$

Where $σ = \frac{M}{π R^{2}}$

$I = \frac{M R^{2}}{2} - \frac{3}{32} M R^{2}$

$I = \frac{13}{32} M R^{2}$

Stream:NEETSubject:PhysicsTopic:RotationSubtopic:Calculation of M.I.

⏱ 2mℹ️ Source: PYQ_2016

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